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Simplified LaGrange Point Calculation

  1. Jan 22, 2010 #1
    I am attempting for my own curiosity to find out at what point during a geodesic path from the Earth to the Moon one would reach a gravitationally neutral point.

    This is essentially the L1, but without adjustments for centripetal force of a moving system, and ignoring all other gravitational bodies (i.e. the sun).

    It's one of those back-of-envelope things that I've run out of envelope for. My stumbling block seems to be the math, but I figured this was a better place to ask for guidance since the subject may have something to do with it.

    So basically, I set

    Fearth = G (mme * mearth) / r2earth and Fmoon = G (mme * mmoon) / r2moon to be equal.

    So when I remove G and move everything so that distances are on one side, and masses are on another, I get

    sqrt( mearth / mmoon ) = rearth / rmoon.

    This can't be right. It's supposed to be an inverse square. However, I cannot for the life of me find where this derivation goes wrong. I suspect there is something fundamental with my original concept, or I'm missing some very obvious mathematical issue in my complete ignorance. Which is caused by my never having taken anything higher than high school geometry/trig and no physics. Please feel free to correct me on any and all points, including my terminology and minutiae.

    Thanks in advance!
  2. jcsd
  3. Jan 23, 2010 #2
    Your derivation looks good to me. Perhaps you're confusing rearth with the earth's radius? rearth is the distance from the earth's center to the Lagrangian point. At the Lagrangian point the object will be closer to the moon than to the earth.
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